Wood diaphragm deflection compatibility.
The conventional approach to diaphragm deflection has always been puzzling. Like many other practicing engineers, the author put his reservations aside and calculated as the code demanded.
In a recent webinar on large diaphragm design, the lecturer casually remarked that something is wrong with the diaphragm deflection equation. It was like a side quest hint in an online game. Quest accepted! Thus began a dive into the calculation of diaphragm deflection to understand the equations mandated by the building code.
In all of structural engineering, there are only two principles: Equilibrium of forces (or energy for methods like virtual work) and Compatibility -that is, a single point on a structure has one deflection under load, and all the members that connect to that point must have that same deflection at that point. The principles are simple, but the devil is in the details, and they can be very complex.
The fundamental problem with diaphragm deflection as we calculate it per code is a violation of compatibility. The diaphragm has a single deflection, but we do not force the individual components to have the same deflection.
The calculation of flexible diaphragm deflection is specified in the American Wood Council (AWC) 2015 Special Design Provisions for Wind and Seismic (SDPWS) as:
This equation and approach have been around since ATC 7 was published in 1981. The first term is bending deflection based on a simple beam model using isolated top and bottom chords. The second term combines the apparent shear stiffness of the diaphragm and the nail slip combined. The “split” version of this term is still contained in the 2018 and 2021 International Building Code (IBC), where diaphragm material shear deflection and staple slip deflection are separate terms. The split version for nailed diaphragms appears in the SDPWS Commentary. The third term is a deflection contribution from the slip at chord splices, typically taken as zero for welded steel chords.
From a practical view, the summation of deflections doesn’t make sense. If one considers the diaphragm system as two springs (neglecting the slip term), then this summation would only make sense if the two springs can deflect independently, allowing each spring to only experience its own calculated deflection. This is a “springs in series” approach to calculating overall deflection. However, the diaphragm only has a single deflection, and both components must have the same deflection. This creates interesting consequences if you “close the loop” on the calculations.
To illustrate, take the simple example of a flexible diaphragm with dimensions and loading as shown. The diaphragm will be the same thickness and nailing throughout to make the example close to the assumptions in the equations. The loads are based on 15 pounds per square foot of roof dead load, a total panel height of 30 feet, and a panel weight of 100 pounds per square foot. Using a Cs value of 0.2 for the 100 × 200-foot diaphragm shown in Figure 1, the calculated diaphragm load v is 900 pounds per linear foot.
To determine an area for the chords, use the typical approach of taking the maximum moment and resisting it as a couple in the chords. In this case, we will assume that it is an angle or channel ledger using ASTM A36 steel:
The bending portion of the deflection is calculated using the first term:
The design shear is divided by a φ factor of 0.8 to get a table value of 1125 p ounds per foot (SDPWS 4.2.3). Using SDPWS Table 4.2A, using blocked 15/32˝ ordinary sheathing on 1-1/2-inch framing and 10d nails at 4 inches on center (Case 1, 2, 3 & 4) gives a shear capacity of 1150 pounds per foot and a Ga value of 21 kips per inch. The shear portion of the deflection is calculated using the second term:
Thus, the total deflection, per code, is 1.03 + 2.14 = 3.17 inches. Typically, the design stops here. The engineer compares the calculated deflection to specified limits and moves on. However, what happens when one calculates the required force to make the chords experience this additional deflection?
Re-arranging the first term of the deflection equation and solving for v using the calculated total deflection gives:
This gives a chord force of:
There is a disconnect (literally) between the deflection calculated in the diaphragm and the forces required to generate that deflection. The shear deflection will also require additional force to match the total deflection. This example has a very rigid diaphragm because of the uniform nailing. In a more realistic setting, where the diaphragm capacity and rigidity are reduced across the length of the diaphragm, the apparent chord force would be even larger as the calculated total deflection increases.
One could consider this a relative rigidity problem – that the force to the two springs would be proportioned based on their relative rigidity and sum up the total deflection. This approach suffers because each of the springs must, by statics, carry the forces assigned. Thus, making a few trial runs made it apparent that this was not the correct approach. Examination of beam shear deflection quickly led into a rabbit hole of theories and differential equations that have been developed or reported. The short answer here is that no one understands beam shear deflection completely and that the equation used for diaphragm shear is likely a decent approximation. This result is left as an exercise for the reader. Good luck.
After the unhelpful review of beam shear equations, the author decided to look back at the nominal approach for diaphragms: trusses. The diaphragm idealization is often referred to as similar to a truss. Indeed, the chord force is based on a truss (more or less) with a non-existent web. So, let’s look at an admittedly odd truss to see how well this assumption works out.
The model truss, illustrated in Figure 2, is intended to mimic a diaphragm ratio of 2:1, similar to the example diaphragm. In this case, the truss has a 40-foot span and a 20-foot depth. The truss is composed of 40 panels (one per foot of span) to minimize the edge effects of the modeling. The truss webs are composed of all pin-connected members. There are four cases for this geometry:
| Case | Webs | Chords |
| 1 – soft web, soft chord | HSS2x2x1/8, E = 1800 ksi | HSS2x2x1/8, E = 29000ksi |
| 2 – soft web, stiff chord | HSS2x2x1/8, E = 1800 ksi | HSS6x6x1/2, E = 29000 ksi |
| 3 – stiff web, soft chord | HSS2x2x1/8, E = 29000 ksi | HSS2x2x1/8, E = 29000ksi |
| 4 – stiff web, stiff chord | HSS2x2x1/8, E = 29000 ksi | HSS6x6x1/2, E = 29000 ksi |
Why use wood modulus values for steel sections? To easily see the effect of the web and chord stiffness on total deflection. This truss example is just to examine the stiffness parameters’ effects on deflections. Yes, these webs would buckle, and there would be some serious geometric nonlinearities if they were real. This is a mathematical model for testing an approach. Always keep in mind the Box Admonition:
“All models are wrong; some models are useful.” – Statistician George Box.
| Case | Deflection at “b” | Deflection at “a” | Maximum top chord axial force | Maximum bottom chord axial force |
| Case 1 – soft web, soft chord | 59.61” | 59.51” | 9.848 k | 9.865 k |
| Case 2 – soft web, stiff chord | 59.54” | 59.70” | 9.865 k | 9.832 k |
| Case 3 – stiff web, soft chord | 3.78” | 3.79” | 9.848 k | 9.865 k |
| Case 4 – stiff web, stiff chord | 3.70” | 3.71” | 9.865 k | 9.832 k |
The theoretical axial chord force for this truss geometry is:
The theoretical chord force value is a little larger than the truss models show. The theoretical chord force can be interpreted as the upper limit of a truss with an infinite number of panels or, say, a diaphragm. There is an edge effect as some of the uniform load “bypasses” the truss in the end diagonals. If this were a two-panel truss, the results would be much different and would not approximate a diaphragm. Not only a bad model for our purposes but a useless one.
Note that the top and bottom chords carry very similar force for all cases regardless of their stiffness or the stiffness of the webs. This is consistent with the statics of the geometry. The results also show that changing chord stiffness has no significant deflection component for the overall diaphragm deflection.
The chord force is based on geometry alone. Thus, the maximum chord force is constant for any given load and truss or diaphragm geometry. Since this force is constant across combinations of stiffness, the total axial strain in the chord is also constant. This strain is horizontal, perpendicular to the applied load, as illustrated in Figure 3. No vertical deflection is developed in the chord from the applied load based on statics alone. The vertical deflection comes from the equilibrium of energy by setting internal strain energy equal to the work performed by the external forces.
The chord vertical deflection is calculated by equating the external work to the strain energy in just the chords. Then one calculates the external work energy to match the internal strain energy. This “virtual work” allows one to then determine a shear deflection. Add the two components together and one has the total deflection. This is the rationale behind the current code equation. This also aligns with the results of Timoshenko’s Beam Theory, where shear deformation adds to the overall deflection due to an increase in internal strain energy.
The truss analyses presented here indicate that the original approach does not apply. The two deflection components are interdependent, and the chords have a limited force and strain based on geometry alone.
Figure 4 shows that the top and bottom chord paths must be different because their lengths are different, yet they have the same maximum vertical deflection. This creates a complication for the external work calculation because the distances “traveled” by the external forces are not the same top and bottom.
The results for the truss cases suggest that there is no single equilibrium vertical deflection state for the chord. As long as the curved path of the chord can accommodate the total axial length of the chord, any path will be acceptable, provided there is something to carry the perpendicular component of the axial force that allows the chord to conform to the path as shown in Figure 5.
Essentially, the shear deflection of the webs pushes or pulls the chord out of a horizontal orientation. This movement results in a perpendicular component of the chord force that the webs must resist. This does NOT increase the force or strain in the chord because the ends of chord are not sufficiently restrained (this is NOT a catenary). This component force acts opposite to the shear deflection and slightly reduces the shear strain energy. The deflection of the chord and web is the same. Equilibrium and compatibility are satisfied.
This result also implies that chord slip will have little effect on the ultimate deflection of the diaphragm – it simply changes the final length of the chord. The chord-produced reduction in shear strain will be less. As the chord slip becomes great or the chord becomes very small, the maximum deflection will approach the web-portion (shear) deflection alone.
Great, let’s eliminate chords! Wait, not so fast. Although they contribute little to deflections, they still resist the edge force that would otherwise need to be resisted at the edge of the diaphragm by statics.
What about Timoshenko? A truss or diaphragm is not a beam and is not modeled well by Timoshenko’s equations or approach. It doesn’t work so well with Euler’s model, either – which is the basis of the chord deflection equation.
Interestingly, the metal deck industry already incorporates this approach into their design deflection equation which only has a shear component. Their rationale is that the chord effect is small because metal deck diaphragms are much stiffer than wood diaphragms. Based on this little example, that’s not true – the chord deflection contribution is small, no matter how stiff the diaphragm. Right answer, poor justification. So, we could use just one equation to represent the diaphragm deflection for both wood and steel diaphragms. Wouldn’t that be nice?■